[Kirche]

An allyl radical has greater stability than a tertiary radical.  Why is it that an allyl carbocation is less stable than a tertiary carbocation?

[Mitch]

I would say it is from a Hyperconjugation effect.

[movies]

Isn't it hyperconjugation that supposedly stabilizes radicals too?

[english]

Hyperconjugating sigma bonds do not stabilize radicals to the same effect as they do cations.  The p orbital of a radical is half-filled.  The hyperconjugated bond between the radical and a sigma bond has 2 bonding electrons and 1 antibonding electron.

[Kirche]

Wouldn't that explanation suggest that the allyl radical would be less stable?

[english]

Wouldn't that explanation suggest that the allyl radical would be less stable?

No because a tertiary radical has hyperconjugating bonds alone as a stabilizing effect.  An allyl radical attains much of its stability from the drop in free energy from delocalization, which ultimately gives us a resonance hybrid.

It's kind of difficult to measure such a thing qualitatively.  The energy values associated with each can give a better reasoning. 


Most of these things are like that.  Being able to predict the relative stabilities of intermediates or products by simple deduction alone.

[Kirche]

That makes sense. I'm still having trouble explaining the relative stabilities though.  With the radical, allyl is more stable than tertiary - makes sense.  But with the allyl carbocation its stability is less than a tertiary - in fact comparable to secondary. 

[english]

That makes sense. I'm still having trouble explaining the relative stabilities though.  With the radical, allyl is more stable than tertiary - makes sense.  But with the allyl carbocation its stability is less than a tertiary - in fact comparable to secondary. 

Hyperconjugating sigma bonds do not stabilize radicals to the same effect as they do cations.

[Kirche]

Thank you for your insite.  Would you mind expanding on that.  I'm not clear on how hyperconjugation effects radicals differently from carbocations.

[Mitch]

Make an interaction diagram between an empty p-orbital(carbocation) and a sigma bond and then compare it to the interation diagram of a p-orbital with one electron(radical) with a sigma bond, and then you'll get it.

[movies]

So the stabilization of an alkyl radical is actually from overlap with the neighboring sigma* orbitals, not the sigma bonds?  I never thought of it that way.

Along those lines, if I recall correctly, benzylic radicals can be stabilized by either EWGs or EDGs on the aromatic system.  How does that work?  I would think that if what I said above about the sigma* stabilization is correct, then we would expect that electron accepting functional groups would be capable of stabilizing radicals, so the EWG thing makes sense.  What about EDGs though?  They don't have the same electron accepting ability.

[Mitch]

I've never met a question in organic chemistry that couldn't be rationalized by stereoelectronic effects or, when neccessary, by hyperconjugation.

[tizrhf]

I think you are making a mistake there. The relative stability of the carbocations is this one : benzylic>allylic>tertiary>secondary>primary. The first two carbocation's stabilities is related to pi conjugation (which of course is best in the case of bezylic radicals) while the other alkyl's stability is related to overlap with neighbouring sigma orbitals. The tertiary carbocations have three neighbouring sigma orbitals with which they could overlap and so there is a greater probability of stabilization than for a secondary or primary carbocation. radicals,the same will be very stable if there is some conjugating efect to stabilize them. That is a general rule: any substituent that can conjugate with carbocations, radicals or carbanions will increase these species' stability more than an inductive effect will do, because the + or - can be dissipated over a longer system of atoms (of course the stabilization requires correcting the anomaly of numbers of electrons in the species, that is supplying electrons to carbocations and trying to take away electrons from carbanions). A radical cand be stabilized both by donating and atracting substiuents. An electron-donating substituent will surely increase the energy of the unpaired electron in the radical but will decrease the energy of the lone pair of the electron-donating substituent (2 electrons stabilized and one destabilized, that means an overall stabilization of the system). The unpaired electron, now being of greater energy will occupy a higher energy SOMO (singly occupied molecular orbital) and will be nucleophile: it will try to react with electrophilic sites(these orbitals have higgh-energy electrons so they will be more willing to donate electrons ,thus being nucleophiles instead of accepting electrons and being electrophiles). The same for electron-taking substituents: there is no lone pair involved, just the delocalization of the unpaired electron into a greater system. this lowers the energy of the electron, placing it in a lower-energy SOMO. This way it will react only with an nucleophile, because these SOMO's are more willing to accept an electron than give it away.

I hope i was clear in my explanation. I know that my way of thinking is a little different.  Please comment on my post.